Question Number 4668 by FilupSmith last updated on 20/Feb/16
$${S}=\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\:^{{i}} \sqrt{{n}},\:\:\:{n}\in\mathbb{R} \\ $$$$\mathrm{Can}\:{S}\:\mathrm{be}\:\mathrm{solved}? \\ $$
Answered by Yozzii last updated on 20/Feb/16
$${S}\left({N}\right)=\underset{{i}=\mathrm{1}} {\overset{{N}} {\sum}}{n}^{\mathrm{0}.\mathrm{5}{i}} =\sqrt{{n}}\underset{{i}=\mathrm{1}} {\overset{{N}} {\sum}}\left({n}^{\mathrm{0}.\mathrm{5}} \right)^{{i}−\mathrm{1}} \\ $$$${S}\left({N}\right)=\frac{\sqrt{{n}}\left(\left(\sqrt{{n}}\right)^{{N}} −\mathrm{1}\right)}{\:\sqrt{{n}}−\mathrm{1}} \\ $$$${S}\left({N}\right)=\frac{\left(\sqrt{{n}}\right)^{{N}+\mathrm{1}} −\sqrt{{n}}}{\:\sqrt{{n}}−\mathrm{1}} \\ $$$$\left(\mathrm{1}\right)\:{For}\:{n}\geqslant\mathrm{1}\Rightarrow\sqrt{{n}}\geqslant\mathrm{1}\Rightarrow\left(\sqrt{{n}}\right)^{{N}+\mathrm{1}} \rightarrow\infty\:{as}\: \\ $$$${N}\rightarrow\infty. \\ $$$$\underset{{N}\rightarrow\infty} {\mathrm{lim}}{S}\left({N}\right)=\frac{\left(\sqrt{{n}}\right)^{\infty} −\sqrt{{n}}}{\:\sqrt{{n}}−\mathrm{1}}=\infty. \\ $$$${S}\left(\infty\right)\:{does}\:{not}\:{exist}.\: \\ $$$$ \\ $$$$\left(\mathrm{2}\right){If}\:\mathrm{0}\leqslant{n}<\mathrm{1}\Rightarrow\mathrm{0}\leqslant\sqrt{{n}}<\mathrm{1} \\ $$$$\Rightarrow\left(\sqrt{{n}}\right)^{{N}+\mathrm{1}} \rightarrow\mathrm{0}\:{as}\:{N}\rightarrow\infty. \\ $$$$\therefore\underset{{N}\rightarrow\infty} {\mathrm{lim}}{S}\left({N}\right)=\frac{\left(\sqrt{{n}}\right)^{\infty} −\sqrt{{n}}}{\:\sqrt{{n}}−\mathrm{1}}=\frac{\sqrt{{n}}}{\mathrm{1}−\sqrt{{n}}} \\ $$$${S}\left(\infty\right)\:{exists}. \\ $$$$ \\ $$$$\left(\mathrm{3}\right){For}\:−\mathrm{1}<{n}<\mathrm{0}\:{let}\:{n}=−{a},\:\mathrm{0}<{a}<\mathrm{1}. \\ $$$$\Rightarrow\sqrt{{n}}=\sqrt{−{a}}={i}\sqrt{{a}}=\sqrt{{a}}\left({cos}\mathrm{0}.\mathrm{5}\pi+{isin}\mathrm{0}.\mathrm{5}\right)={e}^{\mathrm{0}.\mathrm{5}{i}\pi} \sqrt{{a}} \\ $$$$\Rightarrow\left(\sqrt{{n}}\right)^{{N}+\mathrm{1}} ={e}^{\mathrm{0}.\mathrm{5}{i}\pi\left({N}+\mathrm{1}\right)} \left(\sqrt{{a}}\right)^{{N}+\mathrm{1}} \\ $$$${Since}\:\mathrm{0}<{a}<\mathrm{1}\Rightarrow\left(\sqrt{{a}}\right)^{{N}+\mathrm{1}} \rightarrow\mathrm{0}\:{as}\:{N}\rightarrow\infty. \\ $$$$\therefore\underset{{N}\rightarrow\infty} {\mathrm{lim}}{S}\left({N}\right)=\frac{\mathrm{0}×{e}^{\infty{i}} −\sqrt{{n}}}{\:\sqrt{{n}}−\mathrm{1}}=\frac{\sqrt{{n}}}{\mathrm{1}−\sqrt{{n}}} \\ $$$${S}\left(\infty\right)=\frac{{i}\sqrt{{a}}}{\mathrm{1}−{i}\sqrt{{a}}}=\frac{{i}\sqrt{{a}}\left(\mathrm{1}+{i}\sqrt{{a}}\right)}{\mathrm{1}+{a}} \\ $$$${S}\left(\infty\right)=\frac{−{a}+{i}\sqrt{{a}}}{\mathrm{1}+{a}}=\frac{{n}+{i}\sqrt{−{n}}}{\mathrm{1}−{n}}.\Rightarrow{S}\left(\infty\right)\in\mathbb{C}. \\ $$$$ \\ $$$$\left(\mathrm{4}\right)\:{For}\:{n}\leqslant−\mathrm{1}\Rightarrow\left(\sqrt{{n}}\right)^{{N}+\mathrm{1}} \rightarrow{e}^{{i}\infty} \:{if}\:{n}=−\mathrm{1}. \\ $$$$\Rightarrow{a}=\mathrm{1}. \\ $$$${S}\left({N}\right)=\frac{\left(\sqrt{{n}}\right)^{{N}+\mathrm{1}} −\sqrt{{n}}}{\:\sqrt{{n}}−\mathrm{1}} \\ $$$$=\frac{\left({i}\sqrt{{a}}\right)^{{N}+\mathrm{1}} −{i}\sqrt{{a}}}{{i}\sqrt{{a}}−\mathrm{1}} \\ $$$$=\frac{{i}\sqrt{{a}}\left(\left({i}\sqrt{{a}}\right)^{{N}} −\mathrm{1}\right)}{{i}\sqrt{{a}}−\mathrm{1}} \\ $$$$=\frac{{i}\sqrt{{a}}\left(\mathrm{1}−{i}\sqrt{{a}}\right)\left(\mathrm{1}−\left({i}\sqrt{{a}}\right)^{{N}} \right)}{\mathrm{1}+{a}} \\ $$$$=\frac{\left({a}+{i}\sqrt{{a}}\right)\left(\mathrm{1}−\left({i}\sqrt{{a}}\right)^{{N}} \right)}{\mathrm{1}+{a}} \\ $$$$=\frac{{a}−{a}\left({i}\sqrt{{a}}\right)^{{N}} +{i}\sqrt{{a}}−\left({i}\sqrt{{a}}\right)^{{N}+\mathrm{1}} }{\mathrm{1}+{a}} \\ $$$$=\frac{{a}+{i}\sqrt{{a}}}{\mathrm{1}+{a}}−\frac{\left({i}\sqrt{{a}}\right)^{{N}} \left({a}+{i}\sqrt{{a}}\right)}{\mathrm{1}+{a}} \\ $$$${S}\left({N}\right)=\frac{{a}+{i}\sqrt{{a}}}{\mathrm{1}+{a}}\left(\mathrm{1}−{a}^{\mathrm{0}.\mathrm{5}{N}} {e}^{\mathrm{0}.\mathrm{5}\pi{iN}} \right) \\ $$$$\underset{{N}\rightarrow\infty} {\mathrm{lim}}{S}\left({N}\right)=\frac{\mathrm{1}+{i}}{\mathrm{2}}\left(\mathrm{1}−{e}^{{i}\infty} \right)\:{which}\:{is}\:{undefined} \\ $$$${since}\:{e}^{{i}\infty} \:{is}\:{indeterminate}. \\ $$$$ \\ $$$$\left(\mathrm{5}\right){If}\:{a}>\mathrm{1}\:{then}\:\underset{{N}\rightarrow\infty} {\mathrm{lim}}{S}\left({N}\right)=\frac{{a}+{i}\sqrt{{a}}}{\mathrm{1}+{a}}\left(\mathrm{1}−\infty{e}^{{i}\infty} \right) \\ $$$$\infty{e}^{{i}\infty} \:{is}\:{undefined}\Rightarrow{S}\left(\infty\right)\:{is}\:{undefined} \\ $$$${for}\:{n}\leqslant−\mathrm{1}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$